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3 years ago

I need to know the answer so I can correct it so I can fully retake my test

Mathematics

2 answers:

torisob [31]3 years ago

7 0

Answer:

D. all real numbers

Step-by-step explanation:

4(x+3)-7greater than or equal to x+3(x+1)

4x+12-7> or equal to x+3x+3

4x+5>or equal to 4x+3

so, by testing

if x=1, is it true that 4x+5 is greater than or equal to 4x+3

4(1)+5=4(1)+3

9=7

hence it true 9 is greater than 7

Option D is collect

marshall27 [118]3 years ago

5 0

Answer:

Step-by-step explanation:

4(x+3) - 7 ≥ x + 3(x + 1)

4x + 12 -7 ≥ x + 3x + 3

5≥3

so this answer tells us that this equation will always be true no matter what number you plug in for x ... so all real numbers works

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RIGHT ANSWER GETS BRAINLIST 50 POINTS

mixas84 [53]

Answer:

6 cubic cm

Step-by-step explanation:

A = 6a^2

A = 6(1)^2

A = 6(1)

A = 6

3 0

2 years ago

Identify the class width, class midpoints, and class boundaries for the given frequency distribution.

Crank

Answer:

a. Class width=4

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

$M=\frac{lower class limit+upper class limit}{2}$

Class

Interval Midpoints

45-48 $\frac{45+48}{2}=46.5$

49-52 $\frac{49+52}{2}=50.5$

53-56 $\frac{53+56}{2}=54.5$

57-60 $\frac{57+60}{2}=58.5$

61-64 $\frac{61+64}{2}=62.5$

65-68 $\frac{65+68}{2}=66.5$

69-72 $\frac{69+72}{2}=70.5$

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0

3 years ago

What is 4180 rounded to the nearest thousand

lina2011 [118]

4180 rounded to the nearest thousand is as much as 4000. It's because 4180 is closer to 4000 than to 5000.

<span>I hope it will help you with rounding in the future :)</span>

6 0

3 years ago

Read 2 more answers

8x over 3 - x over 2 equals -13

Leya [2.2K]

Equation:
x/3 = x/8 + 2/3
---
Multiply thru by 24 to get:
8x = 3x + 16
5x = 16
x = 3 1/5
========

4 0

4 years ago

In an effort to cut costs and improve profits, any US companies have been turning to outsourcing. In fact, according to Purchasi

My name is Ann [436]

Answer:

a) 0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.

b) 90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.

c) 0.23% probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years.

Step-by-step explanation:

For questions a and b, the normal approximation to the binomial is used, while for question c, the central limit theorem is used.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

54% of companies surveyed outsourced some part of their manufacturing process in the past two to three years.

This means that $p = 0.54$

555 of these companies are contacted.

This means that $n = 555$

Mean and standard deviation: Normal approximation to the binomial:

$\mu = E(X) = np = 555*0.54 = 299.7$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{555*0.54*0.46} = 11.74$

a. What is the probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years?

Using continuity correction, this is $P(X \geq 338 - 0.5) = P(X \geq 337.5)$ , which is 1 subtracted by the p-value of Z when X = 337.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{337.5 - 299.7}{11.74}$

$Z = 3.22$

$Z = 3.22$ has a p-value of 0.9994.

1 - 0.9994 = 0.0006

0.0006*100% = 0.06%

0.06% probability that 338 or more companies outsourced some part of their manufacturing process in the past two or three years.

b. What is the probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years?

Using continuity correction, this is $P(X \geq 285 - 0.5) = P(X \geq 284.5)$ , which is 1 subtracted by the p-value of Z when X = 284.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{284.5 - 299.7}{11.74}$

$Z = -1.29$

$Z = -1.29$ has a p-value of 0.0985.

1 - 0.0985 = 0.9015

0.9015*100% = 90.15%

90.15% probability that 285 or more companies outsourced some part of their manufacturing process in the past two or three years.

c. What is the probability that 48% or less of these companies outsourced some part of their manufacturing process in the past two or three years?

Now we use the sampling distribution of the sample proportions, which have:

$\mu = p = 0.54$